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Elementary operation on a determinant results in -

1) Switching two rows or columns causes the determinant to switch sign

2) Adding a multiple of one row to another causes the determinant to remain the same

3) Multiplying a row as a constant results in the determinant scaling by that constant.

But what happens when we perform these operations on a matrix.

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If you perform such operations on matrix $A$, you obtain another matrix $B$ that

  • shares the same row space as $A$,

  • is row equivalent to $A$, and

  • shares the same reduce row echelon form.

$$B=E_n \ldots E_1 A$$

where $E_i$ are elementary matrices.

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You get another matrix! More seriously, every elementary row operation that can be performed on a $m \times n$ matrix has a corresponding elementary square matrix $E \in M_{m \times m}(\mathbb{F})$ such that applying the row operation on an arbitrary matrix $A$ is the same as multiplying $A$ by $E$ from the left, resulting in $EM$. For example, if $m = n = 2$ and the elementary row operation you consider is the one that switches two rows of a $2 \times 2$ matrix, the corresponding $E$ is

$$ E = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$

and we have

$$ EA = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \begin{pmatrix} a_{21} & a_{22} \\ a_{11} & a_{12} \end{pmatrix}. $$

Saying that if you switch two rows, the determinant changes sign is then the statement

$$ \det(EA) = (-1)\det(A) $$

which is generalized by the fact that

$$ \det(EA) = \det(E)\det(A) $$

whether or not $E$ is an elementary matrix. When $E$ describes the row switching operation, we have $\det(E) = -1$. When $E$ describes the multiplication by a scalar $\lambda$ operation, we have $\det(E) = \lambda$. When $E$ describes the addition of a multiple of a row to another row, we have $\det(E) = 1$.

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